ap physics spring and momentum collision question

Question

A 0.100kg block is attached to an initially unstretched spring of force constant k = 50.0 N/m. The block is released from rest at time t = 0.

At the bottom of its oscillation, the block runs into a 0.05 kg clay ball and becomes stuck to it. What will be the new speed of the block at its equilibrium position?

an answer to a previous and relevant question for the same situation was

A 0.100kg block is attached to an initially unstretched spring of force constant k = 50.0 N/m. The block is released from rest at time t = 0.

At what amount of stretch, x, will the block be at equilibrium position?

answer= 0.02 meters

Answer 1

Let

M = 0.1 kg be the mass of the block,

m = 0.05 kg be the mass of the clay ball,

k = 50 N/m be the spring constant,

g = 10 m/s² be gravitational acceleration,

A = 0.02 m be the amplitude of oscillation of the block without the clay ball (see the relevant question),

v (to be computed) be the speed of the block with the clay ball passing through equilibrium position.

At the moment the block gets attached to the clay ball, the spring is extended distance A.

Therefore the spring energy is

E = kA²/2

At the time the block with the clay ball pass through the equilibrium position,

the spring energy is 0, and all the original energy E got converted into kinetic energy

(M+m)v²/2

Therefore

(M+m)v²/2 = kA²/2

v = A√(k/(M+m))

Substituting actual numbers

v = 0.02√(50/0.15) = 0.36 m/s